Problem: You have found the following ages (in years) of all 6 turtles at your local zoo: $ 47,\enspace 24,\enspace 44,\enspace 27,\enspace 28,\enspace 101$ What is the average age of the turtles at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 6 turtles at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{47 + 24 + 44 + 27 + 28 + 101}{{6}} = {45.2\text{ years old}} $ Find the squared deviations from the mean for each turtle. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $47$ years $1.8$ years $3.24$ years $^2$ $24$ years $-21.2$ years $449.44$ years $^2$ $44$ years $-1.2$ years $1.44$ years $^2$ $27$ years $-18.2$ years $331.24$ years $^2$ $28$ years $-17.2$ years $295.84$ years $^2$ $101$ years $55.8$ years $3113.64$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{3.24} + {449.44} + {1.44} + {331.24} + {295.84} + {3113.64}} {{6}} $ $ {\sigma^2} = \dfrac{{4194.84}}{{6}} = {699.14\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{699.14\text{ years}^2}} = {26.4\text{ years}} $ The average turtle at the zoo is 45.2 years old. There is a standard deviation of 26.4 years.